Question: Let $f(x, y) = 2x^2y$ and $g(t) = (\cos(t), \sin(t))$. $h(t) = f(g(t))$ $h'(\pi) = $
Explanation: Formula The multivariable chain rule says that $\dfrac{dh}{dt} = \nabla f(g(t)) \cdot g'(t)$. The $g'(t)$ part is how much a change in $t$ will cause the input to $f$ to move, and the $\nabla f(g(t))$ part is how much $f$ will change in response to this update to its input. [What's the intuition behind the formula?] Applying the formula We want to find $h'(\pi) = \nabla f(g(\pi)) \cdot g'(\pi)$. $\begin{aligned} &g(\pi) = (-1, 0) \\ \\ &g'(\pi) = (-\sin(\pi), \cos(\pi)) = (0, -1) \\ \\ &\nabla f = (4xy, 2x^2) \\ \\ &\nabla f(g(\pi)) = \nabla f(-1, 0) = (0, 2) \end{aligned}$ Substituting: $h'(\pi) = (0, 2) \cdot (0, -1) = -2$ Answer $h'(\pi) = -2$